Inverse of funtion.

1. Oct 5, 2012

trixitium

I'm reading the first chapter of Topology by Munkres. There we can see:

"if $f$ is bijective, there exists a function from B to A called the inverse of $f$.

(...)

As another situation where care is needed, we note that it is not in general true that

$f^{-1}(f(A_0) = A_0$ and $f(f^{-1}(B_0)) = B_0$. The relevant rules, which we leave you to check, are the following: If $f: A \rightarrow B$ and $A_0 \subset A$ and $B_0 \subset B$, then

$A_0 \subset f^{-1}(f(A_0))$ and $f(f^{-1}(B_0) \subset B_0$

The first inclusion is equality if $f$ is injective and the second inclusion is equality if $f$ is surjective."

Are there any sense in talking about inverse considering that $f$ is not injective or surjective???

2. Nov 2, 2012

Latrace

Ah, this is a common misunderstanding, due to a notation issue. If $f : A \rightarrow B$ is any function and $B_0 \subset B$, then by $f^{-1}(B_0)$ people always mean the set $\{ x \in A | f(x) \in B_0 \}$. Notice that this has meaning even if $f$ is not a bijection. Likewise, if $A_0 \subset A$, we have $f(A) = \{ f(x) | x \in A \}$. I think that with these definitions you should be able to understand the conclusions made in your book.

3. Nov 2, 2012

HallsofIvy

Staff Emeritus
Oh, dear, oh, dear! Just seeing this question makes me want to hide under the bed!

The very first time I had to give an explanation of a proof to a class in a topology class, it involved $f^{-1}(A)$ for A a set and I did the whole thing assuming f was invertible!

If f is a function from set X to set Y, and A is a subset of X, B a subset of Y, then we define f(A) to be the set of all y in Y such that f(x)= y for some x in A and $f^{-1}(B)$ to be the set of all x in X such that f(x) is in B.

IF f is "one to one and onto", that is, if f is invertibe, then we can show that $f^{-1}(f(A))= A$, but f does not have to be invertible, or even defined on set B for $f^{-1}(B)$ to be defined.

For example, let f:R=>R be defined by f(x)= x2 and let B= [-4, 4]. Then $f^{-1}(B)= [-2, 2]$. f(2)= f(-2)= 4 so both 2 and -2 are in $f^{-1}(B)$ and for any x between -2 and 2, -4< 0< f(x)< 4, so x is also in $f^{-1}(B)$. If x< -2 or x> 2, f(x)> 4 so not in [-4, 4].

Even $f^{-1}([-4, -1])$ is defined. Because there is NO x such that $f(x)= x^2$ is in [-4, -1] so $f^{-1}([-4, -1])$ is the empty set.